Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $n = \dfrac{x^2 + 9x - 10}{x - 9} \times \dfrac{2x - 18}{x - 1} $
Explanation: First factor the quadratic. $n = \dfrac{(x - 1)(x + 10)}{x - 9} \times \dfrac{2x - 18}{x - 1} $ Then factor out any other terms. $n = \dfrac{(x - 1)(x + 10)}{x - 9} \times \dfrac{2(x - 9)}{x - 1} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ (x - 1)(x + 10) \times 2(x - 9) } { (x - 9) \times (x - 1) } $ $n = \dfrac{ 2(x - 1)(x + 10)(x - 9)}{ (x - 9)(x - 1)} $ Notice that $(x - 9)$ and $(x - 1)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ 2\cancel{(x - 1)}(x + 10)(x - 9)}{ (x - 9)\cancel{(x - 1)}} $ We are dividing by $x - 1$ , so $x - 1 \neq 0$ Therefore, $x \neq 1$ $n = \dfrac{ 2\cancel{(x - 1)}(x + 10)\cancel{(x - 9)}}{ \cancel{(x - 9)}\cancel{(x - 1)}} $ We are dividing by $x - 9$ , so $x - 9 \neq 0$ Therefore, $x \neq 9$ $n = 2(x + 10) ; \space x \neq 1 ; \space x \neq 9 $